Question

6.Знайти розв'язки системи рiвнянь:
x³- y³ = -35
x² + xy + y² = 7

Answer 1

Ответ: (-3;2),  (-2;3).

Объяснение:

$\displaystyle\\\left \{ {{x^3-y^3=-35} \atop {x^2+xy+y^2=7}} \right. \ \ \ \ \left \{ {{(x-y)(x^2+xy+y^2)=-35} \atop {x^2+xy+y^2=7}} \right. \ \ \ \ \left \{ {{(x-y)*7=-35\ |:7} \atop {x^2+xy+y^2-3xy=7-3xy}} \right.\\\\\\\left \{ {{x-y=-5} \atop {x^2-2xy+y^2=7-3xy}} \right. \ \ \ \ \left \{ {{x-y=-5} \atop {(x-y)^2=7-3xy}} \right.\ \ \ \ \left \{ {{x-y=-5} \atop {(-5)^2=7-3xy}} \right.$

$\displaystyle\\\left \{ {{x-y=-5} \atop {25=7-3xy}} \right.\ \ \ \ \left \{ {{x=y-5} \atop {18=-3xy\ |:(-3)}}\ \ \ \ \left \{ {{x=y-5} \atop {xy=-6}} \right. \right. \ \ \ \ \left \{ {{x=y-5} \atop {(y-5)*y=-6}} \right. \\\\\\\left \{ {{x=y-5} \atop {y^2-5y+6=0}} \right. \ \ \ \ \left \{ {{x=y-5} \atop {y^2-2y-3y+6=0}} \right. \ \ \ \ \left \{ {{x=y-5} \atop {y(y-2)-3(y-2)=0}} \right.$

$\displaystyle\\\left \{ {{x=y-5} \atop {(y-2)(y-3)=0}} \right. \ \ \ \ \left \{ {{x_1=-3\ \ \ \ x_2=-2} \atop {y_1=2\ \ \ \ y_2=3}} \right. .$

$\displaystyle\\\left \{ {{x=y-5} \atop {xy=-6}} \right. \ \ \ \ \left \{ {{x=y-5} \atop {(y-5)*y=-6}} \right.$

$\displaystyle\\\left \{ {{x=y-5} \atop {y*(y-2)-3*(y-2)=0}} \right.$